The perpendcular form of the straight line
$y=x+4$ is $x\cos \alpha +y\sin \alpha =p$ then

The correct option is C $p=2\sqrt{2}$
$y=x+4-x+y=\mathrm{4...}\left(i\right)$
Dividing equation $\left(i\right)$ by $\sqrt{(-1{)}^2+1^2}=\sqrt{2}$ ,we get
$\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=2\sqrt{2}\Rightarrow x\cos {135}^{\circ }+y\sin {135}^{\circ }=2\sqrt{2}\Rightarrow \alpha ={135}^{\circ },p=2\sqrt{2}$